What is x if ${log}_{4} x = 2 - {log}_{4} (x + 6)$ $log_4 x=2 - log_4 (x+6)$ ?

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What is x if ${log}_{4} x = 2 - {log}_{4} (x + 6)$ $log_4 x=2 - log_4 (x+6)$ ?

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Answer 1 · 2018-06-08T11:10:50

See process below

Explanation:

In this type of equations, our goal is to arrive to an expresion like
${log}_{b} A = {log}_{b} C$ $log_bA=log_bC$ from this, we conclude $A = C$ $A=C$

Lets see

${log}_{4} x = 2 - {log}_{4} (x + 6)$ $log_4 x=2-log_4(x+6)$

We know that $2 = {log}_{4} 16$ $2=log_4 16$ . then

${log}_{4} x = {log}_{4} 16 - {log}_{4} (x + 6) = {log}_{4} (\frac{16}{x + 6})$ $log_4 x=log_4 16-log_4(x+6)=log_4(16/(x+6))$ applying the rule

$log (\frac{A}{B}) = log A - log B$ $log(A/B)=logA-logB$

So, we have $x = \frac{16}{x + 6}$ $x=16/(x+6)$

$x^{2} + 6 x - 16 = 0$ $x^2+6x-16=0$
by quadratic formula

$x = \frac{- 6 \pm \sqrt{36 + 64}}{2} = \frac{- 6 \pm 10}{2}$ $x=(-6+-sqrt(36+64))/2=(-6+-10)/2$

Solutions are $x_{1} = - 8$ $x_1=-8$ and $x_{2} = 2$ $x_2=2$ we reject negative and the only valid solution is $x_{2}$ $x_2$

Answer 2 · 2018-06-08T12:49:26

$x = 2$ $x=2$

Explanation:

$using the laws of logarithms$ $"using the "color(blue)"laws of logarithms"$

$∙ x log x + log y = log (x y)$ $•color(white)(x)logx+logy=log(xy)$

$∙ x {log}_{b} x = n \Leftrightarrow x = b^{n}$ $•color(white)(x)log_b x=nhArrx=b^n$

$add {log}_{4} (x + 6) to both sides$ $" add "log_4(x+6)" to both sides"$

${log}_{4} x + {log}_{4} (x + 6) = 2$ $log_4x+log_4(x+6)=2$

${log}_{4} x (x + 6) = 2$ $log_4x(x+6)=2$

$x (x + 6) = 4^{2} = 16$ $x(x+6)=4^2=16$

$x^{2} + 6 x - 16 = 0$ $x^2+6x-16=0$

$(x + 8) (x - 2) = 0$ $(x+8)(x-2)=0$

$x = - 8 or x = 2$ $x=-8" or "x=2$

$x > 0 and x + 6 > 0$ $x>0" and "x+6>0$

$thus x = - 8 is invalid$ $"thus "x=-8" is invalid"$

$\Rightarrow x = 2 is the solution$ $rArrx=2" is the solution"$

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What is x if ${log}_{4} x = 2 - {log}_{4} (x + 6)$ $log_4 x=2 - log_4 (x+6)$ ?

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What is x if ${log}_{4} x = 2 - {log}_{4} (x + 6)$ $log_4 x=2 - log_4 (x+6)$ ?